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Elanso [62]
3 years ago
13

To test the effects of a new fertilizer, 100 plots were divided in half. Fertilizer A is randomly applied to one half, and B to

the other. This is ;
(a) an observational study.
(b) a matched pairs experiment.
(c) a completely randomized experiment.
(d) a block design, but not a matched pairs experiment.
(e) impossible to classify unless more details of the study are provided
Engineering
2 answers:
Kruka [31]3 years ago
7 0

Answer:

(b) a matched pairs experiment.

Explanation:

Matched pairs experiment are a special case of a random block design and is usually used when the experiment only has two conditions. In this case the conditions are the different fertilizers.

Oduvanchick [21]3 years ago
3 0

Answer: (b) a matched pairs experiment.

Explanation:

A match pair design or experiment is a randomized experimental procedure. It can be used in the experiment with two different sets of conditions and the subjects can be grouped into pairs. It is based on some of the blocking variable. The subjects are randomly assigned to different treatments.  

The given situation is an example of a matched pair experiment. This is because of the fact that there are two sets of conditions that is the use of fertilizer A and fertilizer B on different halves of the plot randomly to determine their effects.

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A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar
Westkost [7]

Answer:

a) m_{2} = 0.753\,kg, b) Q_{in} = 2122.963\,kJ

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.02726\,\frac{m^{3}}{kg}

u = 1192.94\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.2

State 2 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.06643\,\frac{m^{3}}{kg}

u = 1718.12\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.5

The model for the rigid tank is created by using the First Law of Thermodynamics:

Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}

Initial and final masses are:

m_{1} = \frac{V_{1}}{\nu_{1}}

m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }

m_{1} = 1.834\,kg

m_{2} = \frac{V_{2}}{\nu_{2}}

m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }

m_{2} = 0.753\,kg

a) The final mass within the tank is:

m_{2} = 0.753\,kg

b) The total amount of heat transfer is:

Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}

Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )

Q_{in} = 2122.963\,kJ

5 0
3 years ago
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